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Arc Length and Curvature · Tutorial 178 of 999

Arc Length of a Graph

Learn why a graph’s length is the integral of the square root of one plus the square of its derivative, and how to compute that length in several cases.

Intermediate 10 min read

What You'll Learn

  • Derive the arc-length formula for a graph on a closed interval
  • Identify the differentiability conditions that make the formula applicable
  • Simplify the integrand when the derivative has a useful algebraic form
  • Evaluate an arc length when the integrand does not simplify to a polynomial
  • Check calculations by differentiating an antiderivative and inspecting the integrand
  • Avoid confusing the derivative with its square in the length formula

Introduction

A graph’s length is not generally the horizontal distance between its endpoints. As the graph rises and falls, its path is longer than the interval it spans along the \(x\)-axis. The derivative \(f'(x)\) measures the graph’s local slope, so it determines how much extra length accumulates.

For a sufficiently smooth function \(y=f(x)\), the length of the graph from \(x=a\) to \(x=b\) is found by integrating a quantity built from that slope. The resulting formula can be applied directly, but its derivation also explains why the derivative appears squared and why the horizontal contribution is \(1\).

The Arc-Length Formula for a Graph

Theorem: If \(f\) is continuously differentiable on the closed interval \([a,b]\), where \(a\le b\), then the length \(L\) of the graph \(y=f(x)\) from \(x=a\) to \(x=b\) is
$$ L=\int_a^b \sqrt{1+\bigl(f'(x)\bigr)^2}\,dx. $$

The condition that \(f'\) is continuous ensures that the graph is smooth enough for this integral to describe its length. The integrand is always at least \(1\), since \(\bigl(f'(x)\bigr)^2\ge 0\). Thus the length is at least \(b-a\), the horizontal distance between the endpoints.

The formula gives the length of the portion of the graph whose \(x\)-coordinate runs from \(a\) to \(b\). It does not require the graph to be increasing or decreasing throughout the interval. If the derivative changes sign, its square still contributes correctly to the length.

Why the Derivative Determines Length

Take a partition \(a=x_0<x_1<\cdots<x_n=b\) of the interval. Over one small subinterval \([x_{i-1},x_i]\), the graph’s endpoints are \((x_{i-1},f(x_{i-1}))\) and \((x_i,f(x_i))\). The straight-line distance between them is

$$ \sqrt{(\Delta x_i)^2+(\Delta y_i)^2} = \Delta x_i\sqrt{1+\left(\frac{\Delta y_i}{\Delta x_i}\right)^2}, $$

where \(\Delta x_i=x_i-x_{i-1}>0\) and \(\Delta y_i=f(x_i)-f(x_{i-1})\). By the Mean Value Theorem, there is a point \(c_i\) in \((x_{i-1},x_i)\) such that \(\frac{\Delta y_i}{\Delta x_i}=f'(c_i)\). Consequently, that chord length equals

$$ \Delta x_i\sqrt{1+\bigl(f'(c_i)\bigr)^2}. $$

Adding these chord lengths gives a Riemann sum for the function \(\sqrt{1+(f'(x))^2}\). Since \(f'\) is continuous, this function is continuous too. As the partition is refined, the chord sums approach the length of the graph, while the corresponding Riemann sums approach the integral. This yields the graph-length formula.

The expression under the square root comes directly from the Pythagorean theorem: one leg is the horizontal change \(\Delta x_i\), and the other is the vertical change \(\Delta y_i\). After factoring out \(\Delta x_i\), the ratio \(\Delta y_i/\Delta x_i\) becomes the local slope. That is why the formula contains \(1+\bigl(f'(x)\bigr)^2\), not just the derivative itself.

Worked Example: A Derivative That Makes the Integrand a Polynomial

Worked Example: A Derivative That Makes the Integrand a Polynomial

Find the length of the graph \(f(x)=\frac{2}{3}(x^2+1)^{3/2}\) on \([0,1]\).

First differentiate using the chain rule:

$$ f'(x) = \frac{2}{3}\cdot\frac{3}{2}(x^2+1)^{1/2}\cdot 2x = 2x\sqrt{x^2+1}. $$

Substitute this derivative into the arc-length formula:

$$ \begin{aligned} L &=\int_0^1\sqrt{1+\left(2x\sqrt{x^2+1}\right)^2}\,dx\\ &=\int_0^1\sqrt{1+4x^2(x^2+1)}\,dx\\ &=\int_0^1\sqrt{4x^4+4x^2+1}\,dx. \end{aligned} $$

The expression under the square root is a perfect square:

$$ 4x^4+4x^2+1=(2x^2+1)^2. $$

Because \(2x^2+1>0\) for every real \(x\), its square root is \(2x^2+1\). Therefore,

$$ \begin{aligned} L &=\int_0^1(2x^2+1)\,dx\\ &=\left[\frac{2}{3}x^3+x\right]_0^1\\ &=\left(\frac{2}{3}+1\right)-0\\ &=\frac{5}{3}. \end{aligned} $$

To check, the derivative calculation gave \(1+(f'(x))^2=4x^4+4x^2+1=(2x^2+1)^2\). Thus the simplified integrand is correct, and evaluating its antiderivative at the stated endpoints gives the length \(\frac{5}{3}\).

Worked Example: A Length Integrand That Does Not Become a Polynomial

Worked Example: A Length Integrand That Does Not Become a Polynomial

Find the length of \(f(x)=x^2\) on \([0,1]\).

Differentiate and form the integrand:

$$ f'(x)=2x, \qquad \sqrt{1+\bigl(f'(x)\bigr)^2}=\sqrt{1+4x^2}. $$

This integrand does not simplify to a polynomial. An antiderivative is

$$ F(x)=\frac{x}{2}\sqrt{1+4x^2} +\frac{1}{4}\operatorname{asinh}(2x), $$

where \(\operatorname{asinh}(u)=\ln\!\left(u+\sqrt{u^2+1}\right)\). To check the antiderivative, differentiate its two terms:

$$ \begin{aligned} \frac{d}{dx}\left(\frac{x}{2}\sqrt{1+4x^2}\right) &=\frac{1}{2}\sqrt{1+4x^2} +\frac{2x^2}{\sqrt{1+4x^2}},\\ \frac{d}{dx}\left(\frac{1}{4}\operatorname{asinh}(2x)\right) &=\frac{1}{2\sqrt{1+4x^2}}. \end{aligned} $$

Adding these derivatives and using a common denominator gives

$$ \begin{aligned} F'(x) &=\frac{\frac12(1+4x^2)+2x^2+\frac12}{\sqrt{1+4x^2}}\\ &=\frac{1+4x^2}{\sqrt{1+4x^2}}\\ &=\sqrt{1+4x^2}. \end{aligned} $$

Now evaluate at the endpoints. Since \(\operatorname{asinh}(0)=0\),

$$ \begin{aligned} L &=\int_0^1\sqrt{1+4x^2}\,dx\\ &=F(1)-F(0)\\ &=\frac{\sqrt{5}}{2}+\frac{1}{4}\operatorname{asinh}(2)\\ &=\frac{\sqrt{5}}{2}+\frac{1}{4}\ln(2+\sqrt{5}). \end{aligned} $$

Numerically, this is approximately \(1.4789\). The exact expression is the length; the decimal is only an approximation.

Worked Example: A Simplification That Works on Both Sides of Zero

Worked Example: A Simplification That Works on Both Sides of Zero

Find the length of \(f(x)=\frac{1}{3}(x^2+2)^{3/2}\) on \([-1,2]\).

Differentiate:

$$ f'(x) = \frac{1}{3}\cdot\frac{3}{2}(x^2+2)^{1/2}\cdot 2x = x\sqrt{x^2+2}. $$

Then the length integrand becomes

$$ \begin{aligned} \sqrt{1+\bigl(f'(x)\bigr)^2} &=\sqrt{1+x^2(x^2+2)}\\ &=\sqrt{x^4+2x^2+1}\\ &=\sqrt{(x^2+1)^2}\\ &=x^2+1. \end{aligned} $$

The final equality holds throughout \([-1,2]\), including negative \(x\), because \(x^2+1\) is always positive. Integrate and evaluate:

$$ \begin{aligned} L &=\int_{-1}^{2}(x^2+1)\,dx\\ &=\left[\frac{x^3}{3}+x\right]_{-1}^{2}\\ &=\left(\frac{8}{3}+2\right)-\left(-\frac{1}{3}-1\right)\\ &=\frac{14}{3}-\left(-\frac{4}{3}\right)\\ &=6. \end{aligned} $$

The integrand check is \(1+x^2(x^2+2)=(x^2+1)^2\), with positive square root \(x^2+1\). Its antiderivative differentiates back to \(x^2+1\), and the endpoint calculation gives length \(6\).

Conditions and Practical Choices

Before applying the formula, identify the interval and check the function’s regularity there. A continuously differentiable function on \([a,b]\) meets the conditions stated in the theorem. If a graph is described by different continuously differentiable formulas on finitely many subintervals, compute the length on each piece and add the results, provided the pieces join to form the graph being measured.

The integral may or may not have an elementary antiderivative. When it does not simplify, the formula still defines the length; the integral may need to be estimated numerically. The tutorial on non-elementary arc-length integrals treats such estimates. Here the key point is to set up the correct integrand before deciding how to evaluate it.

If the graph is measured in physical units, those units are retained by the length. For example, if both axes represent distances in metres, then \(f'(x)\) is a ratio of lengths and is dimensionless; integrating the dimensionless square-root factor with respect to \(x\), measured in metres, gives a length in metres.

Common Mistake: Not Squaring the Derivative

A frequent error is to use \(\sqrt{1+f'(x)}\) in place of \(\sqrt{1+(f'(x))^2}\). For \(f(x)=x^2\), this incorrect setup gives \(\sqrt{1+2x}\). At \(x=1\), it produces the factor \(\sqrt{3}\). But the slope there is \(f'(1)=2\), so the correct local length factor is

$$ \sqrt{1+\bigl(f'(1)\bigr)^2} = \sqrt{1+2^2} = \sqrt{5}. $$

The correction is to square the derivative before adding \(1\). The formula comes from the Pythagorean theorem applied to horizontal and vertical changes, and the vertical change contributes its square.

Interpreting the Result

At a point where the graph is horizontal, \(f'(x)=0\), so the local length factor is \(1\): a small horizontal change contributes approximately its horizontal size. Where the slope has larger magnitude, \(\sqrt{1+(f'(x))^2}\) is larger, reflecting the additional distance traveled along the graph. A negative slope increases length in exactly the same way as a positive slope of equal magnitude, because the formula uses the square.

Key takeaway: To find the length of a smooth graph on \([a,b]\), differentiate \(f\), square the derivative, add \(1\), take the square root, and integrate from \(a\) to \(b\). Check any algebraic simplification, especially when taking the square root of a perfect square.

Check Your Understanding

For each question, begin by identifying the derivative and the interval before evaluating or simplifying the length integral.

  1. State the arc-length formula for \(y=f(x)\) on \([a,b]\), and give a sufficient regularity condition for using it.
  2. Set up, but do not evaluate, the arc-length integral for \(f(x)=3x\) on \([0,2]\).
  3. For \(f(x)=\frac{1}{2}x^2\) on \([0,1]\), find the integrand and explain why it is at least \(1\).
  4. Why does a negative value of \(f'(x)\) not make the arc-length integrand smaller than \(1\)?
  5. For the graph \(f(x)=x^2\) on \([0,1]\), identify the incorrect step in replacing the length integrand with \(\sqrt{1+2x}\).

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